Note that Triangle ECK similar to ∆DGK ( case AA)We have CF/FD=CE/DG=CK/KGSo KG//DG and ∆CKF similar to ∆CGD ( case SAS)From ∆CKF similar to ∆CGD we haveDG/FK=5/x=CG/CK=(CK+KG)/CK= 1+KG/CK= 7/2So x=10/7